How to compute the probability distribution of $w=r_1\cdot a+r_2$ in $\mathbb{F}^*_p$ Hypothesis: Let $\mathbb{F}^*_p$ be a finite field, where $p$ is a prime number. $r_1,r_2$ are elements of the field picked uniformly at random. Let $\alpha$ be a fixed element of the field. 
Question 1: How to compute the distribution of $w=r_1\cdot \alpha+r_2$?
Question 2: How to compute the distribution of $w=r_1\cdot \alpha+r_2$, when $w\neq 0$?

Please feel free to answer one of the qustions.
 A: First, if $r_1, r_2$ are randomly chosen from $\mathbb{F}^*_p$, and $\alpha$ is a fixed element of $\mathbb{F}^*_p$.
Question 1:
$x = r_2 \cdot \alpha$ is a uniformly random element of the multiplicative group. $w$ is the sum of two independent uniformly random elements of the multiplicative group, which is not uniformly distributed.
The probability of $w = 0$ is $\frac{1}{p-1}$, while the probability of other field elements is $\frac{p-2}{(p-1)^2}$.
(The reason is that for any $x$ there is an $r_2$ such that $x+r_2 = 0$. However, for any $y \neq 0$ there is the possibility that $x=y$ which means $x + r_2 \neq y$ regardless of $r_2$.)
Question 2:
The answer depends on how $w \neq 0$ is "enforced", but if it is done with rejection sampling (generate new $r_1$, $r_2$ if $w = 0$), it is a uniformly random element of the multiplicative group. I.e. uniformly random out of the non-zero elements of the field.

Second, if $r_1, r_2$ are random elements of the $\mathbb{F}_p$ field. 
Question 1:
Regardless of $\alpha$ and $r_1$, $r_1 \cdot \alpha$ is some element of the field that does not depend on $r_2$. Since $r_2$ is picked uniformly at random, $w$ is uniformly random as well.
Question 2:
The answer again depends on how $w \neq 0$ is "enforced", but with rejection sampling the answer is the same as above.
